# Not Principal Ideal

• Mar 30th 2011, 11:03 PM
gummy_ratz
Not Principal Ideal
How would you prove it's not a P.I.D. by using I?

Let http://www.math.mtu.edu/graduate/comp/img81.gif be an ideal of http://www.math.mtu.edu/graduate/comp/img82.gif. Prove that I is not a principal ideal of http://www.math.mtu.edu/graduate/comp/img82.gif.

I've been trying this for a long time... and I'm just saying okay it has this form, and then I multiply it by (a+bsqrt(-5)) and this happens... but I keep getting so many terms, that I can't find a contradiction. Please help!
• Mar 31st 2011, 01:03 AM
Swlabr
Quote:

Originally Posted by gummy_ratz
How would you prove it's not a P.I.D. by using I?

Let http://www.math.mtu.edu/graduate/comp/img81.gif be an ideal of http://www.math.mtu.edu/graduate/comp/img82.gif. Prove that I is not a principal ideal of http://www.math.mtu.edu/graduate/comp/img82.gif.

I've been trying this for a long time... and I'm just saying okay it has this form, and then I multiply it by (a+bsqrt(-5)) and this happens... but I keep getting so many terms, that I can't find a contradiction. Please help!

Think about norms. $N(2)=4$, $N(1+\sqrt{-5})=1+25 = 26$ (in general, $N(a+b\sqrt{-5})=a^2+25b^2$).

You should therefore notice that if I is a principle ideal, the principle element (or what ever it is called - lets denote it by x) has norm 2 (why?). You should also notice that there exists some a and b such that $x(a+b\sqrt{-5}) = 1+\sqrt{-5}$ and $N(a+b\sqrt{-5})=13$. So...can you find a contradiction in what I have just written?
• Mar 31st 2011, 07:21 AM
gummy_ratz
hmm okay so, 2 is in I. So x*(a + b*sqrt(-5)) = 2. So N(x*a)*N(b*sqrt(-5)) = N(2)
So x^2*a^2 + x^2*25b^2 = 4. But 1 is not in I. So b=0, a=1, x=2.

But then 1 + sqrt(-5) is in I. So x^2*a^2 + x^2*25b^2 = 26. But we know x=2, so 4a^2 + 100b^2 =26, which is not possible.

Is that what you mean? Also, could you create a norm like this for any ideal you were considering in Z[sqrt(-5)]?
• Mar 31st 2011, 08:02 AM
Swlabr
Quote:

Originally Posted by gummy_ratz
hmm okay so, 2 is in I. So x*(a + b*sqrt(-5)) = 2. So N(x*a)*N(b*sqrt(-5)) = N(2)
So x^2*a^2 + x^2*25b^2 = 4. But 1 is not in I. So b=0, a=1, x=2.

But then 1 + sqrt(-5) is in I. So x^2*a^2 + x^2*25b^2 = 26. But we know x=2, so 4a^2 + 100b^2 =26, which is not possible.

Is that what you mean? Also, could you create a norm like this for any ideal you were considering in Z[sqrt(-5)]?

Not quite...have you come across norms before? The point is that $N(xy)=N(x)N(y)$ for all x, y in your ring. So, specifically, $N(x(a+b\sqrt{-5})=N(x)N(a+b\sqrt{-5})$. Use this.
• Mar 31st 2011, 10:16 AM
gummy_ratz
Why would the norm be http://www.mathhelpforum.com/math-he...88b99cba7b.png ?

Wouldn't it be a^2 + 5b^2?

Since (a+bsqrt(-5))*(a-bsqrt(-5)) = a^2 + 5b^2?
• Mar 31st 2011, 01:10 PM
tonio
Quote:

Originally Posted by gummy_ratz
How would you prove it's not a P.I.D. by using I?

Let http://www.math.mtu.edu/graduate/comp/img81.gif be an ideal of http://www.math.mtu.edu/graduate/comp/img82.gif. Prove that I is not a principal ideal of http://www.math.mtu.edu/graduate/comp/img82.gif.

I've been trying this for a long time... and I'm just saying okay it has this form, and then I multiply it by (a+bsqrt(-5)) and this happens... but I keep getting so many terms, that I can't find a contradiction. Please help!

Using, perhaps, the idea of norm Swlabr wrote (and you're right, it must be 5 and not 25. A

typo, surely) , prove that $6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$ are two fundamentally

different factorizations of 6 in that ring and, thus, there is not unique factorization there, so

it cannot be a PID.

Tonio
• Apr 1st 2011, 05:17 AM
Swlabr
Quote:

Originally Posted by tonio
Using, perhaps, the idea of norm Swlabr wrote (and you're right, it must be 5 and not 25. A

typo, surely) , prove that $6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$ are two fundamentally

different factorizations of 6 in that ring and, thus, there is not unique factorization there, so

it cannot be a PID.

Tonio

Yes, well, more of a mental-o (I was just squaring everything in sight!)